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Determinants
96
Lecture 12
Lecture 12
Properties of the Determinant
12
...
Notice that
det A = aj1 aj2
If we let cj = Cj1 Cj2 · · · Cjn then
Cj1
Cj2
· · · ajn
...
Cjn
det A = aj · cTj
...
The following theorem describes how the determinant behaves
under elementary row operations of Type 1
...
1: Suppose that A ∈ Rn×n and let B be the matrix obtained by interchanging two rows of A
...
a21 a22
a11 a12
...
Consider the 2 × 2 case
...
The general case is proved by induction
...
97
Properties of the Determinant
Corollary 12
...
If A ∈ Rn×n has two rows (or two columns) that are equal then
Proof
...
Let B be the matrix obtained by
interchanging rows j and k
...
But clearly
B = A, and therefore det B = det A
...
Now we consider how the determinant behaves under elementary row operations of Type
2
...
3: Let A ∈ Rn×n and let B be the matrix obtained by multiplying a row of
A by β
...
Proof
...
The rows of A
and B different from j are equal, and therefore
Bjk = Ajk ,
for k = 1, 2,
...
In particular, the (j, k) cofactors of A and B are equal
...
Then,
expanding det B along the jth row:
det B = (βaj ) · cTj
= β(aj · cTj )
= β det A
...
Theorem 12
...
Then det B = det A
...
For any matrix A and any row vector r = [r1 r2 · · · rn ] the expression
r · cTj = r1 Cj1 + r2 Cj2 + · · · + rn Cjn
is the determinant of the matrix obtained from A by replacing the jth row with the row r
...
The jth row of B is bj = aj + βak
...
Example 12
...
Suppose that A is a 4 × 4 matrix and suppose that det A = 11
...
Interchanging (or swapping) rows changes the sign of the determinant
...
Example 12
...
Suppose that A is a 4 × 4 matrix and suppose that det A = 11
...
If B is obtained from A by replacing row a3 by 3a1 + a3 ,
what is det B?
Solution
...
Therefore,
det B = 11
...
7
...
Let
a1 , a2 , a3 , a4 denote the rows of A
...
This is not quite a Type 3 elementary row operation because a3 is multiplied by
7
...
Therefore, expanding det B along the third row
det B = (3a1 + 7a3 ) · cT3
= 3a1 · cT3 + 7a3 · cT3
= 7(a3 · cT3 )
= 7 det A
= 77
99
Properties of the Determinant
Example 12
...
Suppose that A is a 4 × 4 matrix and suppose that det A = 11
...
If B is obtained from A by replacing row a3 by 4a1 + 5a2 ,
what is det B?
Solution
...
The third row of B is
b3 = 4a1 + 5a2
...
2
Determinants and Invertibility of Matrices
The following theorem characterizes invertibility of matrices with the determinant
...
9: A square matrix A is invertible if and only if det A 6= 0
...
Beginning with the matrix A, perform elementary row operations and generate a
sequence of matrices A1 , A2 ,
...
Thus, matrix Ai is obtained from Ai−1 by performing one of the elementary row operations
...
1, 12
...
4, if det Ai−1 6= 0 then det Ai 6= 0
...
Now, Ap is triangular and therefore its determinant is the product
of its diagonal entries
...
In
this case, A is invertible because there are r = n leading entries in Ap
...
In this case, A is not invertible because there are
r < n leading entries in Ap
...
12
...
Theorem 12
...
Then det B = β n det A
...
Consider the 2 × 2 case:
βa11 βa12
det(βA) =
βa12 βa22
= βa11 · βa22 − βa12 · βa21
= β 2 (a11 a22 − a12 a21 )
= β 2 det A
...
Consider a 3 × 3 matrix A
...
The general case can be treated using mathematical induction on n
...
11
...
What is
det(3A)?
Solution
...
Theorem 12
...
Then
det(AB) = det(A) det(B)
...
13: For any square matrix det(Ak ) = (det A)k
...
14: If A is invertible then
det(A−1 ) =
1
...
From AA−1 = In we have that det(AA−1) = 1
...
Therefore
det(A) det(A−1 ) = 1
or equivalently
det A−1 =
1
...
15
...
Suppose that det A = 3, det B = 0, and
det C = 7
...
(i): We have det(AC) = det A det C = 3 · 7 = 21
...
(ii): We have det(AB) = det A det B = 3 · 0 = 0
...
(iii): We have det(ACB) = det A det C det B = 3·7·0 = 0
...
After this lecture you should know the following:
• how the determinant behaves under elementary row operations
• that A is invertible if and only if det A 6= 0
• that det(AB) = det(A) det(B)
102
Lecture 13
Lecture 13
Applications of the Determinant
13
...
If cj = Cj1 Cj2 · · · Cjn then
Cj1
C
j2
det A = aj1 aj2 · · · ajn
...
...
To compute det B expand along its jth row bj = ak :
det B = ak · cTj = 0
...
Recall that for any matrix A ∈ Rn×n , if we expand along the jth row then
det A = aj · cTj
...
(
det A, if j = k
aj · cTk =
0,
if j 6= k
...
...
...
···
...
· · · Cnn
cn
a1
a2
T
A(Cof(A)) =
...
an
a1 cT1
det A
a1 cT2
· · · a1 cTn
a cT a cT · · · a cT
2 1
2 2
2 n
=
...
...
...
...
T
T
T
an c1 an c2 · · · an cn
This can be written succinctly as
0
=
...
0
0
···
0
0
...
...
· · · det A
det A · · ·
...
0
A(Cof(A))T = det(A)In
...
A
det A
This leads to the following formula for the inverse:
A−1 =
1
(Cof(A))T
det A
Although this is an explicit and elegant formula for A−1, it is computationally intensive,
even for 3 × 3 matrices
...
Indeed, if A =
we have Cof(A) =
and therefore
c d
−b a
A
−1
1
d −b
...
Let us first be clear about what we mean by an integer matrix
...
1: A matrix A ∈ Rm×n is called an integer matrix if every entry of A is
an integer
...
Then det(A) is a non-zero integer
and (Cof(A))T is an integer matrix
...
Now det(A) det(A−1) = 1 thus it must be the case that det(A) = ±1
...
Then by the Cofactor method
A−1 =
1
(Cof(A))T = ±(Cof(A))T
det(A)
and therefore A−1 is also an integer matrix
...
Theorem 13
...
We can use the previous theorem to generate integer matrices with an integer inverse
as follows
...
By construction, det(M0 ) = ±1
...
This generates a sequence of matrices
M1 ,
...
Moreover,
M0 ∼ M1 ∼ · · · ∼ Mp
...
105
Applications of the Determinant
13
...
The formula is known as Cramer’s Rule
...
Using the Cofactor method, A−1 = det1 A (Cof(A))T , and therefore
C11 C21
C
1 12 C22
x=
...
det A
...
C1n C2n
b1
· · · Cn1
b2
· · · Cn2
...
...
bn
· · · Cnn
Consider the first component x1 of x:
x1 =
1
(b1 C11 + b2 C21 + · · · + bn Cn1 )
...
...
...
= b1 C11 + b2 C21 + · · · + bn Cn1
...
· · · ann
Similarly,
x2 =
1
(b1 C12 + b2 C22 + · · · + bn Cn2 )
det A
and (b1 C12 + b2 C22 + · · · + bn Cn2 ) is the expansion of the determinant along the second
column of the matrix obtained from A by replacing the second column with b
...
3: (Cramer’s Rule) Let A ∈ Rn×n be an invertible matrix
...
Then the
solution to Ax = b is
det A1
1
det A2
x=
...
det An
Although this is an explicit and elegant formula for x, it is computationally intensive, and
used mainly for theoretical purposes
...
3
Volumes
The volume of the parallelepiped determined by the vectors v1 , v2 , v3 is
Vol(v1 , v2 , v3 ) = abs(v1T (v2 × v3 )) = abs(det v1 v2 v3 )
where abs(x) denotes the absolute value of the number x
...
How are Vol(v1 , v2 , v2 ) and Vol(w1 , w2 , w2 ) related?
Compute:
Vol(w1 , w2 , w3 ) = abs(det w1 w2 w3 )
= abs det Av1 Av2 Av3
= abs det(A v1 v2 v3 )
= abs det A · det v1 v2 v3
= abs(det A) · Vol(v1 , v2 , v3 )
...
In summary:
Theorem 13
...
Let A be the matrix of a linear transformation and let w1 , w2 , w3 be
the images of v1 , v2 , v3 under A, respectively
...
Example 13
...
Consider
4
A= 2
1
the data
1 −1
1
0
−1
4 1 , v1 = −1 , v2 = 1 , v3 = 5
...
Find the volume of the parallelepiped
spanned by the vectors {w1 , w2 , w3 }
...
We compute:
We compute:
Vol(v1 , v2 , v3 ) = abs(det( v1 v2 v3 )) = abs(−7) = 7
det(A) = 55
...
107
Applications of the Determinant
After this lecture you should know the following:
• what the Cofactor Method is
• what Cramer’s Rule is
• the geometric interpretation of the determinant (volume)
108
Lecture 14
Vector Spaces
14
...
Mathematically speaking, a vector is just an element of a
vector space
...
You
have already worked with several types of vector spaces
...
the set Rn ,
2
...
the set of all functions from [a, b] to R, and
4
...
In all of these sets, there is an operation of “addition“ and “multiplication by scalars”
...
Definition 14
...
If u, v, w are in V and if α, β ∈ R are scalars:
(1) The sum u + v is in V
...
(5) For each v there is a vector −v in V such that v + (−v) = 0
...
(closure under scalar multiplication)
(7) α(u + v) = αu + αv
(8) (α + β)v = αv + βv
(9) α(βv) = (αβ)v
(10) 1v = v
It can be shown that 0 · v = 0 for any vector v in V
...
Example 14
...
Let V be the unit disc in R2 :
V = {(x, y) ∈ R2 | x2 + y 2 ≤ 1}
Is V a vector space?
Solution
...
For example, take u = (1, 0) ∈
V and multiply by say α = 2
...
Therefore, property (6) of the
definition of a vector space fails, and consequently the unit disc is not a vector space
...
3
...
Is V a vector space?
Solution
...
For example, u = (1, 1) is a
point in V but 2u = (2, 2) is not
...
For example, both u = (1, 1) and v = (2, 4) are in V but u + v = (3, 5) and (3, 5) is
not a point on the parabola V
...
110
Lecture 14
Example 14
...
Let V be the graph of the function f (x) = 2x:
V = {(x, y) ∈ R2 | y = 2x}
...
We will show that V is a vector space
...
We first note that an arbitrary point in V can be written as u = (x, 2x)
...
Then
u + v = (a + b, 2a + 2b) = (a + b, 2(a + b))
...
Verify that V is closed under scalar multiplication:
αu = α(a, 2a) = (αa, α2a) = (αa, 2(αa))
...
There is a zero vector 0 = (0, 0) in V:
u + 0 = (a, 2a) + (0, 0) = (a, 2a)
...
Therefore, the graph of the function f (x) = 2x is a vector space
...
To emphasize, a vector space is a set
that comes equipped with an operation of addition and scalar multiplication and these two
operations satisfy the list of properties above
...
5
...
, an ∈ R
...
Let u(t) = u0 + u1 t + · · · + un tn and let v(t) = v0 + v1 t + · · · + vn tn be polynomials
in V
...
111
Vector Spaces
Then u + v is a polynomial of degree at most n and thus (u + v) ∈ Pn [t], and therefore this
shows that Pn [t] is closed under addition
...
The 0 vector in Pn [t] is the zero polynomial 0(t) = 0
...
Thus Pn [t] is a vector space
...
6
...
Under the usual operations
of addition of matrices and scalar multiplication, is Mn×m a vector space?
Solution
...
It is clear that the space
Mm×n is closed under these two operations
...
It can be verified that all other properties of the
definition of a vector space also hold
...
Example 14
...
The n-dimensional Euclidean space V = Rn under the usual operations of
addition and scalar multiplication is vector space
...
8
...
Is V a vector space?
14
...
In
this case, we would say that W is a subspace of V
...
Definition 14
...
A subset W of V is called a subspace of V
if it satisfies the following properties:
(1) The zero vector of V is also in W
...
(3) W is closed under scalar multiplication, that is, if u is in W and α is a scalar then
αu is in W
...
10
...
Is W a subspace of V = R2 ?
112
Lecture 14
Solution
...
Let u = (a, 2a) and
v = (b, 2b) be elements of W
...
| {z
| {z }
x
x
Because the x and y components of u + v satisfy y = 2x then u + v is inside in W
...
Let α be any scalar and let u = (a, 2a) be an element of W
...
All three conditions of a subspace are satisfied for
W and therefore W is a subspace of V
...
11
...
Is W a subspace?
Solution
...
Then the sum
u1 + u2 = (x1 + x2 , y1 + y2 ) has components x1 + y1 ≥ 0 and x2 + y2 ≥ 0 and therefore u1 + u2
is in W
...
For example if u = (1, 1)
and α = −1 then αu = (−1, −1) is not in W because the components of αu are clearly not
non-negative
...
12
...
We define the
trace of a matrix A ∈ Mn×n as the sum of its diagonal entries:
tr(A) = a11 + a22 + · · · + ann
...
Is W a subspace of V?
Solution
...
Suppose
that A and B are in W
...
Consider the matrix
C = A + B
...
Now let α be a scalar and let C = αA
...
Thus, tr(C) = 0, that is, C = αA ∈ W, and consequently W is closed under scalar multiplication
...
Example 14
...
Let V = Pn [t] and consider the subset W of V:
W = {u ∈ Pn [t] | u′ (1) = 0}
In other words, W consists of polynomials of degree n in the variable t whose derivative at
t = 1 is zero
...
The zero polynomial 0(t) = 0 clearly has derivative at t = 1 equal to zero, that is,
0′ (1) = 0, and thus the zero polynomial is in W
...
Then, u′ (1) = 0 and also v′ (1) = 0
...
From the rules of differentiation, we compute
(u + v)′ (1) = u′ (1) + v′ (1) = 0 + 0
...
Now let α
be any scalar and let u(t) be a polynomial in W
...
To determine whether or
not the scalar multiple αu(t) is in W we must determine if αu(t) has a derivative of zero at
t = 1
...
Therefore, the polynomial (αu)(t) is in W and thus W is closed under scalar multiplication
...
Example 14
...
Let V = Pn [t] and consider the subset W of V:
W = {u ∈ Pn [t] | u(2) = −1}
In other words, W consists of polynomia